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ra1l [238]
2 years ago
10

The world’s two highest dams are both in Tajikistan. The Rogun dam is 35 meters taller than the Nurek dam. Together they are 635

meters tall. Find the height of the Nurek dam
Mathematics
1 answer:
Thepotemich [5.8K]2 years ago
7 0

Answer: Rogun dam is 335m tall and Nurek dam is 300m tall

Step-by-step explanation:

I will use r to represent the height of the Rogun dam and n to represent the height of the Nurek dam.

The system of equations is as follows:

r+n=635

r-n=35

From the second equation, we can rearange it to be:

r=35+n

Then take this new value of r and plug it into our first equation to find n

(35+n)+n=635

35+2n=635

2n=600

n=300

This means that the Nurek dam is 300m tall, and the Rogun dam is 335m tall, found by pluggin the value into our first equation:

r+300=635

r=335

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3 years ago
A hiker is hiking in a valley. The height of the valley is h(x,y)=4x2+y2 where x and y are the east-west and north-south distanc
Ainat [17]

Answer:

A. \frac{\partial{h}}{\partial{t}}=0

Step-by-step explanation:

A. The problems asked for 2 ways to solve it, expanding the equation with the substitution  x(t)=2 cos(t) and y(t)=4 sin(t) to differentiate it . The other way is by chain rule.

Expanding and differentiating:

We start by substituting x(t)=2 cos(t) and y(t)=4 sin(t) in h(x,y)=4x2+y2:

h(x,y)=4x^{2}+y^{2}= 4(2cos(t))^{2}+(4sin(t))^{2}\\h(x,y)=4(4cos^{2}(t))+(16sen^{2}(t))\\h(x,y)=16cos^{2}(t)+16sen^{2}(t)=16(sen^{2}(t)+cos^{2}(t))\\h(x,y)=16

So, in the path that the hiker chose:

\frac{\partial{h}}{\partial{t}}=0

Chain rule:

We start differentiating h(x,y) using chain rule as follows:

\frac{\partial{h}}{\partial{t}}= \frac{\partial{h}}{\partial{x}}\frac{\partial{x}}{\partial{t}}+\frac{\partial{h}}{\partial{y}}\frac{\partial{y}}{\partial{t}}

Now, it´s easy to find all these derivatives:

\frac{\partial{h}}{\partial{x}}=8x\\\frac{\partial{x}}{\partial{t}}=-2sin(t)\\\frac{\partial{h}}{\partial{y}}=2y\\\frac{\partial{y}}{\partial{t}}=4cos(t)

Now we replace them in the chain rule, with the replacement x=2cos(t) and y=4sin(t) in the x,y that are left and we operate everything:

\frac{\partial{h}}{\partial{t}}= 8x(-2sin(t))+2y(4cos(t)

\frac{\partial{h}}{\partial{t}}= 8(2cos(t))(-2sin(t))+2(4sin(t))(4cos(t)

\frac{\partial{h}}{\partial{t}}= -32cos(t)sin(t)+32sin(t)cos(t)

\frac{\partial{h}}{\partial{t}}= 0

This will be our answer

6 0
3 years ago
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